Count characters in a string whose ASCII values are prime

Given a string S. The task is to count and print the number of characters in the string whose ASCII values are prime.

Examples:

Input: S = “geeksforgeeks”
Output : 3
‘g’, ‘e’ and ‘k’ are the only characters whose ASCII values are prime i.e. 103, 101 and 107 respectively.

Input: S = “abcdefghijklmnopqrstuvwxyz”
Output: 6



Approach: The idea is to generate all primes upto max ASCII value of character of string S using Sieve of Eratosthenes. Now, Iterate the string and get the ASCII value of each character. If the ASCII value is prime then increment the count. Finally, print the count.

Below is the implementation of the above approach:

C++

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// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
#define max_val 257
  
// Function to find prime characters in the string
int PrimeCharacters(string s)
{
  
    // USE SIEVE TO FIND ALL PRIME NUMBERS LESS
    // THAN OR EQUAL TO max_val
    // Create a Boolean array "prime[0..n]". A
    // value in prime[i] will finally be false
    // if i is Not a prime, else true.
    vector<bool> prime(max_val + 1, true);
  
    // 0 and 1 are not primes
    prime[0] = false;
    prime[1] = false;
    for (int p = 2; p * p <= max_val; p++) {
  
        // If prime[p] is not changed, then
        // it is a prime
        if (prime[p] == true) {
  
            // Update all multiples of p
            for (int i = p * 2; i <= max_val; i += p)
                prime[i] = false;
        }
    }
  
    int count = 0;
  
    // Traverse all the characters
    for (int i = 0; i < s.length(); ++i) {
        if (prime[int(s[i])])
            count++;
    }
  
    return count;
}
  
// Driver program
int main()
{
    string S = "geeksforgeeks";
  
    // print required answer
    cout << PrimeCharacters(S);
  
    return 0;
}

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Java

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// Java implementation of above approach 
class Solution
{
static final int max_val=257
  
// Function to find prime characters in the String 
static int PrimeCharacters(String s) 
  
    // USE SIEVE TO FIND ALL PRIME NUMBERS LESS 
    // THAN OR EQUAL TO max_val 
    // Create a Boolean array "prime[0..n]". A 
    // value in prime[i] will finally be false 
    // if i is Not a prime, else true. 
    boolean prime[]= new boolean[max_val+1];
      
    //initilize the value
    for(int i=0;i<=max_val;i++)
    prime[i]=true;
  
    // 0 and 1 are not primes 
    prime[0] = false
    prime[1] = false
    for (int p = 2; p * p <= max_val; p++) { 
  
        // If prime[p] is not changed, then 
        // it is a prime 
        if (prime[p] == true) { 
  
            // Update all multiples of p 
            for (int i = p * 2; i <= max_val; i += p) 
                prime[i] = false
        
    
  
    int count = 0
  
    // Traverse all the characters 
    for (int i = 0; i < s.length(); ++i) { 
        if (prime[(int)(s.charAt(i))]) 
            count++; 
    
  
    return count; 
  
// Driver program 
public static void main(String args[])
    String S = "geeksforgeeks"
  
    // print required answer 
    System.out.print( PrimeCharacters(S)); 
  
}
//contributed by Arnab Kundu

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Python3

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# Python3 implementation of above approach 
  
from math import sqrt
  
max_val = 257 
  
# Function to find prime characters in the string 
def PrimeCharacters(s) :
  
    # USE SIEVE TO FIND ALL PRIME NUMBERS LESS 
    # THAN OR EQUAL TO max_val 
    # Create a Boolean array "prime[0..n]". A 
    # value in prime[i] will finally be false 
    # if i is Not a prime, else true.
    prime = [True] * (max_val + 1)
  
    # 0 and 1 are not primes 
    prime[0] = False
    prime[1] = False 
    for p in range(2, int(sqrt(max_val)) + 1) :
  
        # If prime[p] is not changed, then 
        # it is a prime 
        if (prime[p] == True) : 
  
            # Update all multiples of p 
            for i in range(2*p ,max_val + 1, p) :
                prime[i] = False
  
    count = 0
  
    # Traverse all the characters 
    for i in range(len(s)) :
        if (prime[ord(s[i])]) :
            count += 1
              
    return count 
  
# Driver program 
if __name__ == "__main__"
  
    S = "geeksforgeeks"
  
    # print required answer 
    print(PrimeCharacters(S)) 
  
# This code is contributed by Ryuga

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C#

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// C# implementation of above approach 
using System;
class GFG{
      
static readonly int max_val = 257; 
  
// Function to find prime characters in the String 
static int PrimeCharacters(String s) 
    // USE SIEVE TO FIND ALL PRIME NUMBERS LESS 
    // THAN OR EQUAL TO max_val 
    // Create a Boolean array "prime[0..n]". A 
    // value in prime[i] will finally be false 
    // if i is Not a prime, else true. 
    bool []prime = new bool[max_val + 1]; 
  
    //initilize the value 
    for(int i = 0; i <= max_val; i++) 
    prime[i] = true
  
    // 0 and 1 are not primes 
    prime[0] = false
    prime[1] = false
    for (int p = 2; p * p <= max_val; p++) 
    
        // If prime[p] is not changed, then 
        // it is a prime 
        if (prime[p] == true
        
            // Update all multiples of p 
            for (int i = p * 2; i <= max_val; i += p) 
            prime[i] = false
              
        
          
    
      
    int count = 0; 
      
    // Traverse all the characters 
    for (int i = 0; i < s.Length; ++i) 
    
        if (prime[(int)(s[i])]) 
        count++; 
          
    
    return count; 
      
  
// Driver Code
public static void Main() 
    String S = "geeksforgeeks"
      
    // print required answer 
    Console.Write( PrimeCharacters(S)); 
      
  
// This code is contributed by PrinciRaj1992

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Output:

8


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